Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $r = \dfrac{n - 1}{-3n^2 + 12n} \div \dfrac{-4n^2 - 20n + 24}{n^3 + n^2 - 30n} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{n - 1}{-3n^2 + 12n} \times \dfrac{n^3 + n^2 - 30n}{-4n^2 - 20n + 24} $ First factor out any common factors. $r = \dfrac{n - 1}{-3n(n - 4)} \times \dfrac{n(n^2 + n - 30)}{-4(n^2 + 5n - 6)} $ Then factor the quadratic expressions. $r = \dfrac {n - 1} {-3n(n - 4)} \times \dfrac {n(n + 6)(n - 5)} {-4(n + 6)(n - 1)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {(n - 1) \times n(n + 6)(n - 5) } {-3n(n - 4) \times -4(n + 6)(n - 1) } $ $r = \dfrac {n(n + 6)(n - 5)(n - 1)} {12n(n + 6)(n - 1)(n - 4)} $ Notice that $(n + 6)$ and $(n - 1)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {n\cancel{(n + 6)}(n - 5)(n - 1)} {12n\cancel{(n + 6)}(n - 1)(n - 4)} $ We are dividing by $n + 6$ , so $n + 6 \neq 0$ Therefore, $n \neq -6$ $r = \dfrac {n\cancel{(n + 6)}(n - 5)\cancel{(n - 1)}} {12n\cancel{(n + 6)}\cancel{(n - 1)}(n - 4)} $ We are dividing by $n - 1$ , so $n - 1 \neq 0$ Therefore, $n \neq 1$ $r = \dfrac {n(n - 5)} {12n(n - 4)} $ $ r = \dfrac{n - 5}{12(n - 4)}; n \neq -6; n \neq 1 $